1. **State the problem:** We have four quadratic functions: $y = A x^2$, $y = B x^2$, $y = C x^2$, and $y = D x^2$. (a) Determine the sign (positive or negative) of each coefficient $A$, $B$, $C$, and $D$ based on the parabola's concavity. (b) Identify which coefficient is closest to zero. (c) Identify which coefficient has the least value. 2. **Recall the rule for parabola concavity:** - If the coefficient of $x^2$ is positive, the parabola opens upward (concave up). - If the coefficient of $x^2$ is negative, the parabola opens downward (concave down). 3. **Apply the rule to each parabola:** - $y = A x^2$ opens downward, so $A$ is negative. - $y = B x^2$ opens downward, so $B$ is negative. - $y = C x^2$ opens upward, so $C$ is positive. - $y = D x^2$ opens upward, so $D$ is positive. 4. **Choose the coefficient closest to zero:** - Among $A$, $B$, $C$, and $D$, $D$ is selected as closest to zero. 5. **Choose the coefficient with the least value:** - Since $A$ and $B$ are negative and $C$ and $D$ are positive, the least value is the most negative. - $B$ is selected as the coefficient with the least value. **Final answers:** - (a) $A$ and $B$ are negative; $C$ and $D$ are positive. - (b) $D$ is closest to zero. - (c) $B$ has the least value.